Question

The Basic buffer solution is made with \(0.1\ \text{M}\ \text{NH}_{4}\text{OH}\) and \(0.2\ \text{M}\ \text{NH}_{4}\text{Cl}\). If \(K_{b}\) for the base is \(1.8\times 10^{-5}\), what is the pH of the buffer?

• A. \(4.74\)
• B. \(9.26\)
• C. \(8.96\)
• D. \(5.05\)

Hint:

Use pOH = pKb + log([salt]/[base]) and then pH = 14 - pOH.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a basic buffer, \(\text{pOH}=\text{p}K_{b}+\log\dfrac{[\text{salt}]}{[\text{base}]}\).

Step 2: Find pKb
\(\text{p}K_{b}=-\log(1.8\times 10^{-5})=4.74\)

Step 3: Calculation
\(\text{pOH}=4.74+\log\dfrac{0.2}{0.1}=4.74+0.30=5.04\)

Step 4: pH
\(\text{pH}=14-5.04=8.96\)

Final Answer:
The pH of the buffer is 8.96. \[ \boxed{\text{(C) }8.96} \]